Question

The area in the first quadrant enclosed by the axis, the line $x=y\sqrt{3}$ and the circle $x^2+y^2=4$ is

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is B $\frac{\pi }{3}$



$x^2+y^2=4\Rightarrow (y\sqrt{3}{)}^2+y^2=4\Rightarrow 3y^2+y^2=4\Rightarrow y^2=1\Rightarrow y=1$
If $y=1$ then $x=\sqrt{3}$
$\therefore$ Required area = ${\int }_0^{\sqrt{3}}\frac{x}{\sqrt{3}}dx+{\int }_{\sqrt{3}}^2\sqrt{4-x^2}dx$
$={\left[\frac{x^2}{2\sqrt{3}}\right]}_0^{\sqrt{3}}+{[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}si{n}^{-1}\frac{x}{2}]}_0^{\sqrt{3}}=\frac{\sqrt{3}}{2}+\pi -\frac{\sqrt{3}}{2}-\frac{2\pi }{3}=\frac{\pi }{3}$ sq.unit